Coprime Solutions to Divisibility Relation

There is one solution, namely $(m,n)=(3,1)$ . Let $(x,y)=\gcd(x,y)$ for simplicity. To show this assume $(m,n)=1$ and the condition holds. We have that $m^3+n^3\equiv (m+n)(m^2+n^2-mn)\equiv -(m+n)(m+n+nm)\mod m^2+n^2+m+n$ So the criteria is equivalent to $\frac{(m+n)(m+n+nm)}{m^2+n^2+m+n}\in\mathbb{Z}\qquad (1)$ From here we have $(m^2+n^2+m+n, m+n)=((m+n)^2-2mn,m+n)=(m+n, 2mn)$ Going further using $(m,n)=1$ we have that $(m+n, 2mn)=(m+n,2)$ . So $(1)$ holds iff $\frac{(m+n,2)\cdot (m+n+mn)}{m^2+n^2+m+n}\in\mathbb{Z}$ Thus $m^2+n^2+m+n\le (m+n,2)\cdot (m+n+mn)$ . If $(m+n,2)=1$ , the inequality doesn't hold. Thus $(m+n,2)=2$ and $m^2+n^2+m+n\mid 2(m+n+nm)$ Suppose $m^2+n^2+m+n\neq 2(m+n+nm)$ , then $m^2+n^2+m+n\mid 2(m+n+nm)/p$ for some prime $p$ . But this would imply $m^2+n^2+m+n\le 2(m+n+nm)/p \le m+n+nm$ which is false. Hence $m^2+n^2+m+n= 2(m+n+nm)$ and $(m-n)^2=m+n$ . This gives $m-n\mid n+m$ , so, also using $m\ge n$ , $m-n=(m-n, n+m)=(m+n,2m)=(m+n,2)=2$ where the third equality comes from the Euclidean Algorithm, with the rest coming from $(m,n)=1$ and the previous deduction about $(m+n,2)=2$ . Finally we have $m=n+2$ , which plugging into $(m-n)^2=n+m$ gives $(m,n)=(3,1)$ , which we verify to satisfy the original criterion.